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Question-28288




Question Number 28288 by Mr eaay last updated on 23/Jan/18
Answered by Rasheed.Sindhi last updated on 23/Jan/18
^• (2+(√3) )^(−1) =(1/(2+(√3)))×((2−(√3))/(2−(√3)))                        =2−(√3)   ^• (2+(√3))^2 =(2)^2 +2(2)((√3))+((√3) )^2                     =7+4(√x)  ^• (2+(√3))^3 =(2)^3 +((√3))^3 +3(2)^2 ((√3) )+  3(2)((√3) )^2                   =8+3(√3) +12(√3) +18                  =26+15(√3)  The given equation⇒      ( (2+(√3))^3 )^x −5((2+(√3))^2 )^x +6(2+(√3))^x +((2+(√3))^(−1) )^x =5      ( (2+(√3))^x )^3 −5((2+(√3))^x )^2 +6(2+(√3))^x +((2+(√3))^x )^(−1) =5  Let (2+(√3) )^x =y  ⇒ y^3 −5y^2 +6y+y^(−1) =5        y^4 −5y^3 +6y^2 −5y+1=0                  ⋮
(2+3)1=12+3×2323=23(2+3)2=(2)2+2(2)(3)+(3)2=7+4x(2+3)3=(2)3+(3)3+3(2)2(3)+3(2)(3)2=8+33+123+18=26+153Thegivenequation((2+3)3)x5((2+3)2)x+6(2+3)x+((2+3)1)x=5((2+3)x)35((2+3)x)2+6(2+3)x+((2+3)x)1=5Let(2+3)x=yy35y2+6y+y1=5y45y3+6y25y+1=0
Commented by Rasheed.Sindhi last updated on 23/Jan/18
��şίŕ ţħάήķş ғόŕ ςόŕŕέςţίόή!��
Answered by ajfour last updated on 23/Jan/18
x=±1
x=±1

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